Question 149

If $$x+\frac{1}{x}=-2$$ then the value of $$x^{p}+x^{q}$$ is: (Where p is an even number and q is an odd number)

Solution

Given : $$x+\frac{1}{x}=-2$$

=> $$\frac{x^2+1}{x}=-2$$

=> $$x^2+1+2x=0$$

=> $$(x+1)^2=0$$

=> $$x+1=0$$

=> $$x=-1$$

$$\therefore$$ $$x^{p}+x^{q}$$ (let $$p=2$$ and $$q=1$$)

=> $$(-1)^2+(-1)^1=1-1=0$$

=> Ans - (D)

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